Please show all work for rating A 0.2-kg particle is moving with a velocity v1,

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A 0.2-kg particle is moving with a velocity v1, = i + j + 2k m/s at time t1, = 1 s. If the single force F = (5 + 3l)i + (2 - I2)j + 3k N acts on the particle, determine its velocity v2 at time t2= 4 s.

Explanation / Answer

F = [(5+3*t) i + (2- t^2)j + 3k] N v1 = (i +j +2k) m/s t1 = 1 s t2 = 4 s a = F/m = F/0.2 => a = [(25+15*t) i + (10 -5*t^2)j + 15k] m/s^2 dv/dt = [(25+15*t) i + (10 -5*t^2)j + 15k] => integral of [dv] from v1 to v2 = integral of [{(25+15*t) i +(10 -5*t^2)j +15k} .dt] from t1 to t2 => v2 - v1 = [(25*t +7.5*t^2)i +(10*t -5*(t^3/3)j + (15*t) k] from t1 to t2 => v2 - (i +j +2k) = [{(25*4 +7.5*4^2)- (25*1 +7.5*1^2)} i +{(10*4 -5*(4^3/3)) -(10*1 -5*(1^3/3))j + {(15*4) -(15*1)} k] => v2 - (i +j +2k) = 187.5 i + (-75)j + 45 k => v2 = (187.5 +1) i +(-75 +1)j +(45 +2)k => v2 = 188.5 i - 74 j + 47 k

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